你好，我正在尝试使用ajax帖子从数据库中获取数据，但是我没有正确地获取任何数据。第一列数据被夹住到另一列中（成员名称出现在image字段和info字段中）。图片也分享了请检查。而且在使用ajax获取数据时，datatables也不起作用。帮我解决这个问题。。。
谢谢和问候
<body>
<label>Party</label>
<select id='partydropdown' name='partydropdown' onchange="partyFunction();">
<option>--select a party--</option>
<?php
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
?>
<option value="<?php echo $row["Id"];?>">
<?php echo $row["PartyName"];?>
</option>
<?php }}?>
</select>
<div id="showhide"></div>
</body>
<script type="text/javascript">
function partyFunction(){
debugger;
$("#showhide").empty();
$("#showhide").html('');
$("#showhide").append("<table class='table table-bordered text-center table-responsive' border='1px' id='example'>"+
"<tr>"+
"<th>PartyMemberName</th>"+
"<th>Image</th>"+
"<th>Info</th>"+
"</tr>"+
"<tbody id='partyBody'>"+
"</tbody>"+
"</table>"
);
$postdata = {};
$postdata["Id"]=$("#partydropdown").val();
console.log($("#partydropdown").val());
$.post('test_data.php',$postdata,function (data) {
debugger;
console.log(data);
console.log(data["data"][0].candiateName);
$("#partyBody").empty();
$("#partyBody").html('');
console.log(data["data"]);
console.log(data["data"].length);
for(var i=0; i<data["data"].length; i++){
if(data["data"][i].candiateName != null){
$("#partyBody").append("<tr>"+
"<td id='resdata"+i+"'></td>"+
"<td id='resdata1"+i+"' ></td>"+
"<td id='resdata2"+i+"'></td>"+
"</tr>");
$("#resdata"+i).text(data["data"][i].candiateName);
$("#resdata1"+i).append("<img id='photo"+i+"'>");
$("#resdata2"+i).text(data["data"][i].Background);
$("#photo"+i).attr('src', 'http://aptsvotes.com/wp-content/themes/apts2019/img'+data["data"][i].Photo );
}
}
});
};
</script>下面是test\u data.php代码
<?php
include_once "conn.php";
include_once "voterdbclass.php";
session_start();
$tbl_name2="Parties";
$dbObj = new Database1();
$values1 = array("all");
$querys = "SELECT c1.CandidateName,c1.Photo,c1.Background ,c1.Type FROM aptsv1_votes.Parties p1 LEFT JOIN aptsv1_votes.Candidates c1 ON c1.CurrentPartyId = p1.Id where p1.Id ='" . $_POST['Id'] . "' limit 21";
$res = $dbObj->SelectRecord($tbl_name2,$values1,"","$querys");
$data=array();
$i=0;
while ($rs = $res->fetch_array(MYSQLI_ASSOC)) {
$data[$i]['candiateName']=$rs['CandidateName'];
$data[$i]['Photo']=$rs['Photo'];
$data[$i]['Background']=$rs['Background'];
$i++;
}
$json_array= array(
"data" =>$data
);
echo json_encode($json_array);
?>
2条答案
按热度按时间hivapdat1#
2izufjch2#
您必须将json数据解析到第一个脚本中。当你发布结果时
json_encode，我发现不见了转化为代码
谢谢您！